Fermi level of copper from conductivity

The conductivity gives the free-electron concentration, and the concentration gives the Fermi energy:

$$\sigma = e n \mu_d \;\Rightarrow\; n = \frac{\sigma}{e\mu_d}, \qquad E_{F0} = \frac{h^2}{8m_e}\left(\frac{3n}{\pi}\right)^{2/3}$$

Given: $\sigma = 63.5 \times 10^{6}\ \text{S/m}$, $\mu_d = 43\ \text{cm}^2\text{V}^{-1}\text{s}^{-1} = 43 \times 10^{-4}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$.

Electron concentration

$$n = \frac{63.5 \times 10^{6}}{1.602 \times 10^{-19} \times 43 \times 10^{-4}} = 9.22 \times 10^{28}\ \text{m}^{-3}$$

Fermi level

$$\begin{aligned} E_{F0} &= \frac{(6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31}} \left(\frac{3 \times 9.22 \times 10^{28}}{\pi}\right)^{2/3} \\ &= 6.025 \times 10^{-38} \times 1.979 \times 10^{19} \\ &= 1.19 \times 10^{-18}\ \text{J} = \frac{1.19 \times 10^{-18}}{1.602 \times 10^{-19}} = 7.44\ \text{eV} \end{aligned}$$

The Fermi level lies 7.44 eV above the bottom of the conduction band (close to the accepted 7.0 eV for copper; the difference comes from the slightly high conductivity value given).

Answer: $n = 9.22 \times 10^{28}\ \text{m}^{-3}$, $E_F \approx 1.19 \times 10^{-18}\ \text{J} = 7.44\ \text{eV}$.